Question

# If one root of the quadratic equation $${ ax }^{ 2 }+bx+c=0$$ is equal to the $$n^{th}$$ power of the other, then show that $${ \left( { ac }^{ n } \right) }^{ \dfrac { 1 }{ n+1 } }+{ \left( { a }^{ n }c \right) }^{ \dfrac { 1 }{ n+1 } }+b=0$$

Solution

## Given that $$\beta = \alpha^n$$Product of roots $$=\alpha \beta =\alpha { \alpha }^{ n }={ \alpha }^{ n+1 }=\dfrac { c }{ a }$$$$\Rightarrow \alpha ={ \left( \dfrac { c }{ a } \right) }^{ \dfrac { 1 }{ n+1 } }$$Substituting $$\alpha$$ in $${ ax }^{ 2 }+bx+c=0$$$$a{ \left( \dfrac { c }{ a } \right) }^{ \dfrac { 2 }{ n+1 } }+b{ \left( \dfrac { c }{ a } \right) }^{ \dfrac { 1 }{ n+1 } }+c=0$$$${ \left( { ac }^{ n } \right) }^{ \dfrac { 1 }{ n+1 } }+{ \left( { a }^{ n }c \right) }^{ \dfrac { 1 }{ n+1 } }+b=0$$Ans: 1Mathematics

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